Question

Prove: Let x and y be positive real numbers. If x ≤ y, then √x ≤ √y.

Answer 1

Suppose that $x,y> 0$ and that $x\leq y$. We can subtract $x$ from both sides to obtain $y-x \geq 0$. Recall that $(a+b)(a-b)=a^2-b^2$. Now we can replace $a$ and $b$ in this identity with $\sqrt{y}$ and $\sqrt{x}$ respectively and we get $(\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x})=y-x$. From this it follows that $(\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x}) \geq 0$. Since $\sqrt{y}+\sqrt{x} > 0$, we can divide by it both sides of the inequality without altering its direction and end up with $\sqrt{y}-\sqrt{x} \geq 0$. Now we just need to add $\sqrt{x}$ to both sides and conclude that $\sqrt{y} \geq \sqrt{x}$, which finishes our proof.